> On 11 Feb 2019, at 13:00, agrayson2...@gmail.com wrote:
> 
> 
> 
> On Monday, February 11, 2019 at 2:20:25 AM UTC, agrays...@gmail.com wrote:
> 
> 
> On Tuesday, February 5, 2019 at 8:43:59 PM UTC, agrays...@gmail.com <> wrote:
> 
> 
> On Monday, February 4, 2019 at 8:56:57 AM UTC-7, Bruno Marchal wrote:
> 
>> On 3 Feb 2019, at 00:03, agrays...@gmail.com <> wrote:
>> 
>> 
>> 
>> On Saturday, February 2, 2019 at 2:59:30 PM UTC-7, agrays...@gmail.com 
>> <http://gmail.com/> wrote:
>> 
>> 
>> On Saturday, February 2, 2019 at 1:40:29 AM UTC-7, Bruno Marchal wrote:
>> 
>>> On 1 Feb 2019, at 21:29, agrays...@gmail.com <> wrote:
>>> 
>>> 
>>> 
>>> On Friday, February 1, 2019 at 5:55:30 AM UTC-7, Bruno Marchal wrote:
>>> 
>>>> On 31 Jan 2019, at 21:10, agrays...@gmail.com <> wrote:
>>>> 
>>>> 
>>>> 
>>>> On Thursday, January 31, 2019 at 6:47:12 AM UTC-7, Bruno Marchal wrote:
>>>> 
>>>>> On 31 Jan 2019, at 01:28, agrays...@gmail.com <> wrote:
>>>>> 
>>>>> 
>>>>> 
>>>>> On Wednesday, January 30, 2019 at 2:38:58 PM UTC-7, agrays...@gmail.com 
>>>>> <http://gmail.com/> wrote:
>>>>> 
>>>>> 
>>>>> On Wednesday, January 30, 2019 at 5:16:05 AM UTC-7, Bruno Marchal wrote:
>>>>> 
>>>>>> On 30 Jan 2019, at 02:59, agrays...@gmail.com <> wrote:
>>>>>> 
>>>>>> 
>>>>>> 
>>>>>> On Tuesday, January 29, 2019 at 4:37:34 AM UTC-7, Bruno Marchal wrote:
>>>>>> 
>>>>>>> On 28 Jan 2019, at 22:50, agrays...@gmail.com <> wrote:
>>>>>>> 
>>>>>>> 
>>>>>>> 
>>>>>>> On Friday, January 25, 2019 at 7:33:05 AM UTC-7, Bruno Marchal wrote:
>>>>>>> 
>>>>>>>> On 24 Jan 2019, at 09:29, agrays...@gmail.com <> wrote:
>>>>>>>> 
>>>>>>>> 
>>>>>>>> 
>>>>>>>> On Sunday, January 20, 2019 at 11:54:43 AM UTC, agrays...@gmail.com 
>>>>>>>> <http://gmail.com/> wrote:
>>>>>>>> 
>>>>>>>> 
>>>>>>>> On Sunday, January 20, 2019 at 9:56:17 AM UTC, Bruno Marchal wrote:
>>>>>>>> 
>>>>>>>>> On 18 Jan 2019, at 18:50, agrays...@gmail.com <> wrote:
>>>>>>>>> 
>>>>>>>>> 
>>>>>>>>> 
>>>>>>>>> On Friday, January 18, 2019 at 12:09:58 PM UTC, Bruno Marchal wrote:
>>>>>>>>> 
>>>>>>>>>> On 17 Jan 2019, at 14:48, agrays...@gmail.com <> wrote:
>>>>>>>>>> 
>>>>>>>>>> 
>>>>>>>>>> 
>>>>>>>>>> On Thursday, January 17, 2019 at 12:36:07 PM UTC, Bruno Marchal 
>>>>>>>>>> wrote:
>>>>>>>>>> 
>>>>>>>>>>> On 17 Jan 2019, at 09:33, agrays...@gmail.com <> wrote:
>>>>>>>>>>> 
>>>>>>>>>>> 
>>>>>>>>>>> 
>>>>>>>>>>> On Thursday, January 17, 2019 at 3:58:48 AM UTC, Brent wrote:
>>>>>>>>>>> 
>>>>>>>>>>> 
>>>>>>>>>>> On 1/16/2019 7:25 PM, agrays...@gmail.com <> wrote:
>>>>>>>>>>>> 
>>>>>>>>>>>> 
>>>>>>>>>>>> On Monday, January 14, 2019 at 6:12:43 AM UTC, Brent wrote:
>>>>>>>>>>>> 
>>>>>>>>>>>> 
>>>>>>>>>>>> On 1/13/2019 9:51 PM, agrays...@gmail.com <> wrote:
>>>>>>>>>>>>> This means, to me, that the arbitrary phase angles have 
>>>>>>>>>>>>> absolutely no effect on the resultant interference pattern which 
>>>>>>>>>>>>> is observed. But isn't this what the phase angles are supposed to 
>>>>>>>>>>>>> effect? AG
>>>>>>>>>>>> 
>>>>>>>>>>>> The screen pattern is determined by relative phase angles for the 
>>>>>>>>>>>> different paths that reach the same point on the screen.  The 
>>>>>>>>>>>> relative angles only depend on different path lengths, so the 
>>>>>>>>>>>> overall phase angle is irrelevant.
>>>>>>>>>>>> 
>>>>>>>>>>>> Brent
>>>>>>>>>>>> 
>>>>>>>>>>>> Sure, except there areTWO forms of phase interference in Wave 
>>>>>>>>>>>> Mechanics; the one you refer to above, and another discussed in 
>>>>>>>>>>>> the Stackexchange links I previously posted. In the latter case, 
>>>>>>>>>>>> the wf is expressed as a superposition, say of two states, where 
>>>>>>>>>>>> we consider two cases; a multiplicative complex phase shift is 
>>>>>>>>>>>> included prior to the sum, and different complex phase shifts 
>>>>>>>>>>>> multiplying each component, all of the form e^i (theta). Easy to 
>>>>>>>>>>>> show that interference exists in the latter case, but not the 
>>>>>>>>>>>> former. Now suppose we take the inner product of the wf with the 
>>>>>>>>>>>> ith eigenstate of the superposition, in order to calculate the 
>>>>>>>>>>>> probability of measuring the eigenvalue of the ith eigenstate, 
>>>>>>>>>>>> applying one of the postulates of QM, keeping in mind that each 
>>>>>>>>>>>> eigenstate is multiplied by a DIFFERENT complex phase shift.  If 
>>>>>>>>>>>> we further assume the eigenstates are mutually orthogonal, the 
>>>>>>>>>>>> probability of measuring each eigenvalue does NOT depend on the 
>>>>>>>>>>>> different phase shifts. What happened to the interference 
>>>>>>>>>>>> demonstrated by the Stackexchange links? TIA, AG 
>>>>>>>>>>>> 
>>>>>>>>>>> Your measurement projected it out. It's like measuring which slit 
>>>>>>>>>>> the photon goes through...it eliminates the interference.
>>>>>>>>>>> 
>>>>>>>>>>> Brent
>>>>>>>>>>> 
>>>>>>>>>>> That's what I suspected; that going to an orthogonal basis, I 
>>>>>>>>>>> departed from the examples in Stackexchange where an arbitrary 
>>>>>>>>>>> superposition is used in the analysis of interference. 
>>>>>>>>>>> Nevertheless, isn't it possible to transform from an arbitrary 
>>>>>>>>>>> superposition to one using an orthogonal basis? And aren't all 
>>>>>>>>>>> bases equivalent from a linear algebra pov? If all bases are 
>>>>>>>>>>> equivalent, why would transforming to an orthogonal basis lose 
>>>>>>>>>>> interference, whereas a general superposition does not? TIA, AG
>>>>>>>>>> 
>>>>>>>>>> I don’t understand this. All the bases we have used all the time are 
>>>>>>>>>> supposed to be orthonormal bases. We suppose that the scalar product 
>>>>>>>>>> (e_i e_j) = delta_i_j, when presenting the Born rule, and the 
>>>>>>>>>> quantum formalism.
>>>>>>>>>> 
>>>>>>>>>> Bruno
>>>>>>>>>> 
>>>>>>>>>> Generally, bases in a vector space are NOT orthonormal. 
>>>>>>>>> 
>>>>>>>>> Right. But we can always build an orthonormal base with a decent 
>>>>>>>>> scalar product, like in Hilbert space, 
>>>>>>>>> 
>>>>>>>>> 
>>>>>>>>> 
>>>>>>>>>> For example, in the vector space of vectors in the plane, any pair 
>>>>>>>>>> of non-parallel vectors form a basis. Same for any general 
>>>>>>>>>> superposition of states in QM. HOWEVER, eigenfunctions with distinct 
>>>>>>>>>> eigenvalues ARE orthogonal.
>>>>>>>>> 
>>>>>>>>> Absolutely. And when choosing a non degenerate 
>>>>>>>>> observable/measuring-device, we work in the base of its eigenvectors. 
>>>>>>>>> A superposition is better seen as a sum of some eigenvectors of some 
>>>>>>>>> observable. That is the crazy thing in QM. The same particle can be 
>>>>>>>>> superposed in the state of being here and there. Two different 
>>>>>>>>> positions of one particle can be superposed.
>>>>>>>>> 
>>>>>>>>> This is a common misinterpretation. Just because a wf can be 
>>>>>>>>> expressed in different ways (as a vector in the plane can be 
>>>>>>>>> expressed in uncountably many different bases), doesn't mean a 
>>>>>>>>> particle can exist in different positions in space at the same time. 
>>>>>>>>> AG
>>>>>>>> 
>>>>>>>> It has a non null amplitude of probability of being here and there at 
>>>>>>>> the same time, like having a non null amplitude of probability of 
>>>>>>>> going through each slit in the two slits experience.
>>>>>>>> 
>>>>>>>> If not, you can’t explain the inference patterns, especially in the 
>>>>>>>> photon self-interference.
>>>>>>>> 
>>>>>>>> 
>>>>>>>> 
>>>>>>>> 
>>>>>>>>> 
>>>>>>>>> Using a non orthonormal base makes only things more complex. 
>>>>>>>>>> I posted a link to this proof a few months ago. IIRC, it was on its 
>>>>>>>>>> specifically named thread. AG
>>>>>>>>> 
>>>>>>>>> But all this makes my point. A vector by itself cannot be superposed, 
>>>>>>>>> but can be seen as the superposition of two other vectors, and if 
>>>>>>>>> those are orthonormal, that gives by the Born rule the probability to 
>>>>>>>>> obtain the "Eigen result” corresponding to the measuring apparatus 
>>>>>>>>> with Eigen vectors given by that orthonormal base.
>>>>>>>>> 
>>>>>>>>> I’m still not sure about what you would be missing.
>>>>>>>>> 
>>>>>>>>> You would be missing the interference! Do the math. Calculate the 
>>>>>>>>> probability density of a wf expressed as a superposition of 
>>>>>>>>> orthonormal eigenstates, where each component state has a different 
>>>>>>>>> phase angle. All cross terms cancel out due to orthogonality,
>>>>>>>> 
>>>>>>>> ?  Sin(alpha) up + cos(alpha) down has sin^2(alpha) probability to be 
>>>>>>>> fin up, and cos^2(alpha) probability to be found down, but has 
>>>>>>>> probability one being found in the Sin(alpha) up + cos(alpha) down 
>>>>>>>> state, which would not be the case with a mixture of sin^2(alpha) 
>>>>>>>> proportion of up with cos^2(alpha) down particles.
>>>>>>>> Si, I don’t see what we would loss the interference terms.
>>>>>>>> 
>>>>>>>> 
>>>>>>>> 
>>>>>>>>> and the probability density does not depend on the phase differences. 
>>>>>>>>>  What you get seems to be the classical probability density. AG 
>>>>>>>> 
>>>>>>>> 
>>>>>>>> I miss something here. I don’t understand your argument. It seems to 
>>>>>>>> contradict basic QM (the Born rule). 
>>>>>>>> 
>>>>>>>> Suppose we want to calculate the probability density of a 
>>>>>>>> superposition consisting of orthonormal eigenfunctions,
>>>>>>> 
>>>>>>> Distinct eigenvalue correspond to orthonormal vector, so I tend to 
>>>>>>> always superpose only orthonormal functions, related to those 
>>>>>>> eigenvalue. 
>>>>>>> 
>>>>>>> 
>>>>>>> 
>>>>>>> 
>>>>>>> 
>>>>>>>> each multiplied by some amplitude and some arbitrary phase shift.
>>>>>>> 
>>>>>>> like  (a up + b down), but of course we need a^2 + b^2 = 1. You need to 
>>>>>>> be sure that you have normalised the superposition to be able to apply 
>>>>>>> the Born rule.
>>>>>>> 
>>>>>>> 
>>>>>>> 
>>>>>>> 
>>>>>>>> If we take the norm squared using Born's Rule, don't all the cross 
>>>>>>>> terms zero out due to orthonormality?
>>>>>>> 
>>>>>>> ?
>>>>>>> 
>>>>>>> The Born rule tell you that you will find up with probability a^2, and 
>>>>>>> down with probability b^2
>>>>>>> 
>>>>>>> 
>>>>>>> 
>>>>>>>> Aren't we just left with the SUM OF NORM SQUARES of each component of 
>>>>>>>> the superposition? YES or NO?
>>>>>>> 
>>>>>>> If you measure in the base (a up + b down, a up -b down). In that case 
>>>>>>> you get the probability 1 for the state above.
>>>>>>> 
>>>>>>> 
>>>>>>> 
>>>>>>>> If YES, the resultant probability density doesn't depend on any of the 
>>>>>>>> phase angles. AG
>>>>>>>> 
>>>>>>>> YES or NO? AG 
>>>>>>> 
>>>>>>> 
>>>>>>> Yes, if you measure if the state is a up + b down or a up - b down.
>>>>>>> No, if you measure the if the state is just up or down
>>>>>>> 
>>>>>>> Bruno
>>>>>>> 
>>>>>>> I assume orthNORMAL eigenfunctions. I assume the probability densities 
>>>>>>> sum to unity. Then, using Born's rule, I have shown that multiplying 
>>>>>>> each component by e^i(theta) where theta is arbitrarily different for 
>>>>>>> each component, disappears when the probability density is calculated, 
>>>>>>> due to orthonormality.
>>>>>> 
>>>>>> 
>>>>>> That seems to violate elementary quantum mechanics. If e^I(theta) is 
>>>>>> different for each components, Born rule have to give different 
>>>>>> probabilities for each components---indeed given by the square of 
>>>>>> e^I(theta).
>>>>>> 
>>>>>> The norm squared of e^i(thetai) is unity, except for the cross terms 
>>>>>> which is zero due to orthonormality. AG 
>>>>>> 
>>>>>>> What you've done, if I understand correctly, is measure the probability 
>>>>>>> density using different bases, and getting different values.
>>>>>> 
>>>>>> The value of the relative probabilities do not depend on the choice of 
>>>>>> the base used to describe the wave. Only of the base corresponding to 
>>>>>> what you decide to measure. 
>>>>>> 
>>>>>> 
>>>>>> 
>>>>>>> This cannot be correct since the probability density is an objective 
>>>>>>> value, and doesn't depend on which basis is chosen. AG
>>>>>> 
>>>>>> Just do the math. Or read textbook.
>>>>>> 
>>>>>> Why don't YOU do the math ! It's really simple. Just take the norm 
>>>>>> squared of a superposition of component eigenfunctions, each multiplied 
>>>>>> by a probability amplitude, and see what you get !  No need to multiply 
>>>>>> each component by e^i(thetai).  Each amplitude has a phase angle 
>>>>>> implied. This is Born's rule and the result doesn't depend on phase 
>>>>>> angles, contracting what Bruce wrote IIUC. If you would just do the 
>>>>>> simple calculation you will see what I am referring to! AG
>>>>> 
>>>>> 
>>>>> Bruce is right. Let us do the computation in the simple case where 
>>>>> e^i(theta) = -1. (Theta = Pi)
>>>>> 
>>>>> Take the superposition (up - down), conveniently renormalised. If I 
>>>>> multiply the whole wave (up - down) by (-1), that changes really nothing. 
>>>>> But if I multiply only the second term, I get the orthogonal state up + 
>>>>> down, which changes everything. (up +down) is orthogonal to (up - down).
>>>>> 
>>>>> Bruno
>>>>> 
>>>>>  Fuck it. You refuse to do the simple math to show me exactly where I 
>>>>> have made an error,  IF I have made an error.  You talk a lot about 
>>>>> Born's rule but I seriously doubt you know how to use  it for simple 
>>>>> superposition. AG 
>>>>> 
>>>>> If you take the inner product squared (Born's rule) using an orthonormal 
>>>>> set of eigenfunctions, you get a sum of the form (a_j)^2 + (b_j) ^2  
>>>>> where A_j is the complex probability amplitude for the jth component, A_j 
>>>>> = a_j + i * b_j. The cross terms drop out due to orthonormality, and the 
>>>>> phase angles are implicitly determined by the relative values of a_j and 
>>>>> b_j for each j.
>>>> 
>>>> If you have prepared the state, so that you know that the state of your 
>>>> object is given by 
>>>> 
>>>>  phi = A_1 up + A_2 down, say, then, if you decide to measure the up/down 
>>>> state, and use the device doing that, you do not need to make the inner 
>>>> product between phi and phi, but between the base state up and/or down to 
>>>> get the probability given by the square of phi * up (to get the 
>>>> probability of up) and the square of phi*down, to get the probability of 
>>>> down. They will both depend on the value of A_1 and A_2. They are 
>>>> respectively (A_1)^2 and (A_2)^2. Of course, we suppose that we have 
>>>> renormalised the state so that (A_1)^2 + (A_2)^2 = 1 (which makes them 
>>>> into probability of getting up and down).
>>>> 
>>>> 
>>>> 
>>>> 
>>>> 
>>>> 
>>>> 
>>>>> The question then becomes how do we calculate the probability density 
>>>>> with the phase angles undetermined.  Are we assuming they are known given 
>>>>> the way the system is prepared? AG
>>>> 
>>>> 
>>>> Yes. The Born rule, written simply, is only that if phi = A_1 up + A_2 
>>>> down, (so the state has been prepared in advance) then if you measure if 
>>>> the object is in up or down, you will find up with a probability given 
>>>> respectively by (A_1)^2 and (A_2)^2.
>>>> All probabilities are relative to the state of the object and the choice 
>>>> of what you decide to measure. It is always simpler to write the state in 
>>>> the base corresponding to the measurement, so that the “simple” Born rule 
>>>> above can be applied immediately.
>>>> 
>>>> Bruno
>>>> 
>>>> For reference I repeat my last comment and add a significant point:
>>>> 
>>>>  If you take the inner product squared (Born's rule) using an orthonormal 
>>>> set of eigenfunctions, you get a sum of the form (a_j)^2 + (b_j) ^2  where 
>>>> A_j is the complex probability amplitude for the jth component, A_j = a_j 
>>>> + i * b_j. The cross terms drop out due to orthonormality, and the phase 
>>>> angles are implicitly determined by the relative values of a_j and b_j for 
>>>> each j. The question then becomes how do we calculate the probability 
>>>> density with the phase angles undetermined.  Are we assuming they are 
>>>> known given the way the system is prepared? AG
>>>> 
>>>> The question for me is how the phase angles are related to interference.
>>> 
>>> But that is explained by may calculation above. You calculation does not 
>>> make sense to me. You compute an inner product of the wave to itself? I 
>>> don’t see the relation with your problem. 
>>> 
>>> Obviously, you don't know how to apply the rule you speak so highly of, 
>>> Born's rule. To calculate the probability density of wf function psi, you 
>>> must calculate <psi, psi>.  Do you dispute this?
>> 
>> Yes, you need to put some projection operator (corresponding to some 
>> eigenvalue you intend to measure) in between. 
>> <psi,psi> is the amplitude of probability to go from the psi state to the 
>> psi state, and should be equal to one (psi being normalised).
>> 
>> Wrong! Not equal to one. I see you like to talk the talk, but refuse to walk 
>> the walk. Just read the 4th paragraph of the Wiki link. I correctly 
>> calculated the probability density for orthonormal eigenstates! AG  
> 
> I understand that if you have a plane wave, say, psi(x,t) = Ae^i(kx-wt), the 
> probability density if given by the square of the square of the modulus of A. 
> I am not sure why you need this, and why this should be problematic with 
> Bruce’s (or mine) explanation.
> 
> I started with a more general case; namely, writing psi as a superposition of 
> orthonormal states, each with its own implied phase shift incorpororated in 
> different A's,. and found that the resultant probability density didn't 
> depend on any cross terms, which zero out due to the orthogonality. 
> (Actually, I started with an explicitly different phase shift of the form 
> e^i(theta), but later omitted that since the phase shifts can be assumed as 
> different and incorporated in the A's.) I thought interference depended on 
> the result for cross terms, but as Phil pointed out, that's not the case. Why 
> I thought this follows from something Feynman wrote in his lectures. I will 
> explain this in another post. Thanks for your help. AG
> 
> To reiterate and clarify; you got your result by taking the inner product of 
> psi with itself, which is what I did, but starting with a more general 
> initial state; namely, a superposition of orthonormal states, each multiplied 
> by a presumably different probability amplitude A_j. Now, if you go back to 
> my earlier comments, you will see the final result; namely, a sum of terms of 
> the form || A_j ||^2, where the cross terms drop out due to orthonormality. 
> So what's my problem? Simply this; somewhere in Feynman's Lectures, he wants 
> to show how quantum probabilities differ from classical probabilities. He 
> shows the difference is between taking the classical probability for say the 
> EM double slit experiment as ||A||^2 + ||B||^2 (where one adds the intensity 
> squared for each wave), and the quantum calculation, ( ||A + B|| )^2, where 
> one first sums the amplitudes *before* taking the resultant norm squared. Why 
> then do I get the first result for the quantum probability -- which 
> presumably is the classical result -- when the initial quantum psi is taken 
> as a sum of orthonormal states? TIA, AG
> 
> Maybe the mistake I am making is confusing the norm squared of a single wf, 
> with that of two distinguishable wf's, one for each slit.  AG


Yes. I think so.

Let me quickly try to sum up the explanation of the difference of quantum and 
classical probability, and, why not, with the two slits, and using Dirac 
Bra-kets, where < f I i > is the amplitude (of probability) to go from the 
initial state I i > to the final state I f >.


I do a drawing:

         1

s                 x

         2     


x, that is I x > is the state of the particle on the screen at position x. Same 
for the other symbols, where 1 and 2 are the location of the slits, on some 
wall, and s is the source location.


The amplitude of probability for going from s to x is given by the bra-ket < x 
I s >

And < x I s > = < x I 1 > <1 I s > + < x I 2 > <2 I s >.

The corresponding probability are given by the square of the amplitudes.

Now, in classical physics, we can in principle knows the slit through which the 
particle go through, and this would give directly the probability:

P(s, x) = (< x I 1 > <1 I s >)^2  + (< x I 2 > <2 I s >.)^2


But in quantum physics, if we don’t observe by which slits the particle go 
through, the probability is given the square of the total amplitude < x I s > = 
  (< x I 1 > <1 I s > + < x I 2 > <2 I s >)^2, which generates the oscillation 
terms.

Opening the other slit makes it impossible for the particle to go at some place 
it was used to go when only one slit was open, surprisingly enough.

In classical physics, we can sum on all the intermediary unknown path 
probabilities, but with quantum physics, we have to sum all amplitudes, and 
only then take the probability (which gives only our own relative probability 
to be there, assuming non-collapse). That is why the position of the particles 
on the screen depends on all the intermediate path. That is what is exploited 
in Shor's algorithm to factorise numbers, for example. 

Bruno




> 
>>> How the phase angles relate to interference is another issue, which I think 
>>> Phil explained. AG
>>> 
>>>> The calculation above shows that the cross terms drop out due to 
>>>> orthonormality.
>>> 
>>> Do it again, explicitly. Take the simple state phi = A_1 up + A_2 down. Up 
>>> and down are orthonormal,
>>> 
>>> Up and Dn are NOT orhonormal.  AG
>>>  
>>> but phi is not orthonormal with either up or down. If “up” means go to the 
>>> left hole, and “down” is go the right hole, the amplitude A_1 and A_2, if 
>>> not null, will interfere, even if only one photon is sent.The wave go 
>>> through both silts, and interfere constructively along some direction and 
>>> destructively along other direction, making it impossible for that photon 
>>> to lend on those last place, like anyway, by the laws of addition of 
>>> sinus/wave.
>>> 
>>>> But IIUC these are the terms which account for interference.
>>> 
>>> I am not sure what you say here. The interferences comes only from the fact 
>>> that we have a superposition of two orthogonal state, and that 
>>> superposition is a new state, which is not orthogonal to either up or down.
>>> 
>>>> Thus, applying Born's rule to a superposition of states where the 
>>>> components are orthonormal, leaves open the question of interference.
>>> 
>>> That does no make sense. The Born rule just say that if you measure 
>>> (up/down) on phi =  A_1 up + A_2 down, you get up with probability (A_1)^2 
>>> and down with probability (A_2)^2. But if you do any measurement, the state 
>>> beg-have like a wave, and the amplitudes add up, constructively or 
>>> destructively.
>>> 
>>> If you don’t understand that, it means you begin to understand quantum 
>>> mechanics, as nobody understand this, except perhaps the Mechanist 
>>> Philosophers …(which predicts something at least as weird and 
>>> counter-intuitive). 
>>> 
>>>> Bruce wrote that the phase angles are responsible for interference. I 
>>>> doubt that result. Am I mistaken? AG
>>> 
>>> Yes, I’m afford you are. The relative phase (in a superposition) angles are 
>>> responsible for the interference. A global phase angle changes nothing.
>>> 
>>> If I am wrong, it's just because I assumed all interference comes from the 
>>> interactions due to the cross terms -- which cancel out for orthonormal 
>>> component states. Also, I never introduced a global phase angle in my 
>>> calculation. If you would do my calculation, or at least understand it, 
>>> you'd understand Born's rule.  I don't need to read Albert's book to 
>>> understand Born's rule. AG
>> 
>> 
>> Once you say that up and down are not orthonormal, I am not sure you have 
>> studied the QM formalism correctly. Any two distinguishable eigenstates of 
>> any observable are orthogonal (and normalised). 
>> 
>> Right. I was mistaken. AG 
> 
> 
> OK. Good.
> 
> 
> 
>> 
>> I have no clue what you don’t understand in my use of the Born rule. You 
>> definitely need to study Albert’s book, I think.
>> 
>> Wiki shows I correctly calculated the probability density. Also I agree with 
>> Phil, and noted the error I made (not in any calculation, but in 
>> interpretation). Didn't you read it? AG  
> 
> That was not enough clear, sorry.
> The wiki is also rather unclear on the Born Rule. I mean that there are 
> clearer exposition.
> 
> 
> 
>> 
>> In your other post you mention wikipedia. No problem there? Actually you can 
>> see that they do put the projection operator at the right place. You can 
>> help yourself with a dictionary, but books and papers are better.
>> 
>> If you put in the projection operator, you're calculating the probability of 
>> getting some eigenvalue, not the probability density of the position. AG 
> 
> 
> You lost me. I was just explaining why the relative phase does play a role 
> for the probability of finding specific values. Bruce was correct, and I 
> still don’t know if you agree on this or not.
> I am not sure that I understand what is your problem.
> 
> Bruno
> 
> 
> 
> 
> 
> 
> 
>> 
>> You could help yourself by reading plain English. SEE PARAGRAPH 4 OF WIKI 
>> LINK. THEY CALCULATED THE PROBABILITY DENSITY AND DIDN'T PUT IN THE 
>> PROJECTION OPERATOR!  AG
>> 
>> Bruno
>> 
>> 
>> 
>> 
>> 
>>> 
>>> I really wish you to read the first 60 pages of David Albert’s book. Its 
>>> exposition of the functioning of the interferometer is crystal clear. I am 
>>> still not sure if you have a problem with the formalism or with the 
>>> weirdness related to it. Read that piece of explanation by Albert, and if 
>>> you still have problem, we can discuss it, but it would be too long (here 
>>> and now) to do that here. 
>>> 
>>> Bruno
>>> 
>>> 
>>> 
>>> 
>>> 
>>> 
>>>> 
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